作者:陈曦
日期:2012-6-8 10:50:13
环境:[Ubuntu 11.04 Intel-based x64 gcc4.5.2 CodeBlocks10.05 AT&T汇编 Intel汇编]
转载请注明出处
Q: 举个例子吧。
A: 下面的代码的目标是计算1+2的值,最后放到变量temp中,并输出:
#include <stdio.h> #include <string.h> #define PRINT_D(longValue) printf(#longValue" is %ld\n", ((long)longValue)); #define PRINT_STR(str) printf(#str" is %s\n", (str)); static void assemble_func() { int temp; __asm__("mov $1, %eax"); __asm__("mov $2, %ebx"); __asm__("add %ebx, %eax"); // 1 + 2 __asm__("mov %%eax, %0":"=r"(temp)); // mov the value of register eax to the var "temp" PRINT_D(temp) // print temp } int main() { assemble_func(); return 0; }
temp is 3
Q: assemble_func函数的汇编代码形式是什么?
A:
0x08048404 <+0>: push ebp 0x08048405 <+1>: mov ebp,esp 0x08048407 <+3>: push ebx 0x08048408 <+4>: sub esp,0x24 => 0x0804840b <+7>: mov eax,0x1 0x08048410 <+12>: mov ebx,0x2 0x08048415 <+17>: add eax,ebx 0x08048417 <+19>: mov ebx,eax 0x08048419 <+21>: mov DWORD PTR [ebp-0xc],ebx 0x0804841c <+24>: mov eax,0x8048510 0x08048421 <+29>: mov edx,DWORD PTR [ebp-0xc] 0x08048424 <+32>: mov DWORD PTR [esp+0x4],edx 0x08048428 <+36>: mov DWORD PTR [esp],eax 0x0804842b <+39>: call 0x8048340 <printf@plt> 0x08048430 <+44>: add esp,0x24 0x08048433 <+47>: pop ebx 0x08048434 <+48>: pop ebp 0x08048435 <+49>: ret
Q: 上面的汇编是内嵌到c代码中的,单独完全的汇编代码,如何实现hello world的功能?
A: 从本质上说,只用汇编的形式需要对于底层更了解,c代码从编译的角度来说和汇编没什么区别,只是写的格式以及调用的东西看起来不一致罢了。
如下,是实现标准控制台输出功能的代码:
.section .rodata str: .ascii "Hello,world.\n" .section .text .globl _main _main: movl $4, %eax # the number of system call movl $1, %ebx # file descriptor, 1 means stdout movl $str, %ecx # string address movl $13, %edx # string length int $0x80
Q: 如何编译它,使用gcc吗?
A: 当然可以,不过这个文件显然不需要预处理了,它已经是汇编格式了,不需要单纯狭义的编译过程了,只需要从汇编过程开始了。
它可以直接生成目标文件hello.o
Q: 接下来做什么?可以直接执行它吗?
A: 试试。
此时,给hello.o添加可执行权限再执行:
Q: 这是为什么?
A: 继续观察hello.o文件的属性。
可以看出,它还不是可执行文件。其实很简单,hello.o只是目标文件,并没有链接成可执行文件。
Q: 这又是为什么?没有找到入口符号_start, ld默认的入口符号是_start?
A: 是的。在代码中使用的是_main, 所以应该让链接器明白,入口符号是_main.
Q: 现在应该可以运行了吧。运行一下:
Hello,world是输出了,为什么后面会出现段错误呢?
A: 我们首先看看上面的运行返回了什么。
返回值为139,它代表什么?
Q: 从系统的errno.h头文件以及相关文件中查找,得到所有系统错误码:
/usr/include/asm-generic/errno-base.h文件:
#ifndef _ASM_GENERIC_ERRNO_BASE_H #define _ASM_GENERIC_ERRNO_BASE_H #define EPERM 1 /* Operation not permitted */ #define ENOENT 2 /* No such file or directory */ #define ESRCH 3 /* No such process */ #define EINTR 4 /* Interrupted system call */ #define EIO 5 /* I/O error */ #define ENXIO 6 /* No such device or address */ #define E2BIG 7 /* Argument list too long */ #define ENOEXEC 8 /* Exec format error */ #define EBADF 9 /* Bad file number */ #define ECHILD 10 /* No child processes */ #define EAGAIN 11 /* Try again */ #define ENOMEM 12 /* Out of memory */ #define EACCES 13 /* Permission denied */ #define EFAULT 14 /* Bad address */ #define ENOTBLK 15 /* Block device required */ #define EBUSY 16 /* Device or resource busy */ #define EEXIST 17 /* File exists */ #define EXDEV 18 /* Cross-device link */ #define ENODEV 19 /* No such device */ #define ENOTDIR 20 /* Not a directory */ #define EISDIR 21 /* Is a directory */ #define EINVAL 22 /* Invalid argument */ #define ENFILE 23 /* File table overflow */ #define EMFILE 24 /* Too many open files */ #define ENOTTY 25 /* Not a typewriter */ #define ETXTBSY 26 /* Text file busy */ #define EFBIG 27 /* File too large */ #define ENOSPC 28 /* No space left on device */ #define ESPIPE 29 /* Illegal seek */ #define EROFS 30 /* Read-only file system */ #define EMLINK 31 /* Too many links */ #define EPIPE 32 /* Broken pipe */ #define EDOM 33 /* Math argument out of domain of func */ #define ERANGE 34 /* Math result not representable */ #endif
/usr/include/asm-generic/errno.h文件:
#ifndef _ASM_GENERIC_ERRNO_H #define _ASM_GENERIC_ERRNO_H #include <asm-generic/errno-base.h> #define EDEADLK 35 /* Resource deadlock would occur */ #define ENAMETOOLONG 36 /* File name too long */ #define ENOLCK 37 /* No record locks available */ #define ENOSYS 38 /* Function not implemented */ #define ENOTEMPTY 39 /* Directory not empty */ #define ELOOP 40 /* Too many symbolic links encountered */ #define EWOULDBLOCK EAGAIN /* Operation would block */ #define ENOMSG 42 /* No message of desired type */ #define EIDRM 43 /* Identifier removed */ #define ECHRNG 44 /* Channel number out of range */ #define EL2NSYNC 45 /* Level 2 not synchronized */ #define EL3HLT 46 /* Level 3 halted */ #define EL3RST 47 /* Level 3 reset */ #define ELNRNG 48 /* Link number out of range */ #define EUNATCH 49 /* Protocol driver not attached */ #define ENOCSI 50 /* No CSI structure available */ #define EL2HLT 51 /* Level 2 halted */ #define EBADE 52 /* Invalid exchange */ #define EBADR 53 /* Invalid request descriptor */ #define EXFULL 54 /* Exchange full */ #define ENOANO 55 /* No anode */ #define EBADRQC 56 /* Invalid request code */ #define EBADSLT 57 /* Invalid slot */ #define EDEADLOCK EDEADLK #define EBFONT 59 /* Bad font file format */ #define ENOSTR 60 /* Device not a stream */ #define ENODATA 61 /* No data available */ #define ETIME 62 /* Timer expired */ #define ENOSR 63 /* Out of streams resources */ #define ENONET 64 /* Machine is not on the network */ #define ENOPKG 65 /* Package not installed */ #define EREMOTE 66 /* Object is remote */ #define ENOLINK 67 /* Link has been severed */ #define EADV 68 /* Advertise error */ #define ESRMNT 69 /* Srmount error */ #define ECOMM 70 /* Communication error on send */ #define EPROTO 71 /* Protocol error */ #define EMULTIHOP 72 /* Multihop attempted */ #define EDOTDOT 73 /* RFS specific error */ #define EBADMSG 74 /* Not a data message */ #define EOVERFLOW 75 /* Value too large for defined data type */ #define ENOTUNIQ 76 /* Name not unique on network */ #define EBADFD 77 /* File descriptor in bad state */ #define EREMCHG 78 /* Remote address changed */ #define ELIBACC 79 /* Can not access a needed shared library */ #define ELIBBAD 80 /* Accessing a corrupted shared library */ #define ELIBSCN 81 /* .lib section in a.out corrupted */ #define ELIBMAX 82 /* Attempting to link in too many shared libraries */ #define ELIBEXEC 83 /* Cannot exec a shared library directly */ #define EILSEQ 84 /* Illegal byte sequence */ #define ERESTART 85 /* Interrupted system call should be restarted */ #define ESTRPIPE 86 /* Streams pipe error */ #define EUSERS 87 /* Too many users */ #define ENOTSOCK 88 /* Socket operation on non-socket */ #define EDESTADDRREQ 89 /* Destination address required */ #define EMSGSIZE 90 /* Message too long */ #define EPROTOTYPE 91 /* Protocol wrong type for socket */ #define ENOPROTOOPT 92 /* Protocol not available */ #define EPROTONOSUPPORT 93 /* Protocol not supported */ #define ESOCKTNOSUPPORT 94 /* Socket type not supported */ #define EOPNOTSUPP 95 /* Operation not supported on transport endpoint */ #define EPFNOSUPPORT 96 /* Protocol family not supported */ #define EAFNOSUPPORT 97 /* Address family not supported by protocol */ #define EADDRINUSE 98 /* Address already in use */ #define EADDRNOTAVAIL 99 /* Cannot assign requested address */ #define ENETDOWN 100 /* Network is down */ #define ENETUNREACH 101 /* Network is unreachable */ #define ENETRESET 102 /* Network dropped connection because of reset */ #define ECONNABORTED 103 /* Software caused connection abort */ #define ECONNRESET 104 /* Connection reset by peer */ #define ENOBUFS 105 /* No buffer space available */ #define EISCONN 106 /* Transport endpoint is already connected */ #define ENOTCONN 107 /* Transport endpoint is not connected */ #define ESHUTDOWN 108 /* Cannot send after transport endpoint shutdown */ #define ETOOMANYREFS 109 /* Too many references: cannot splice */ #define ETIMEDOUT 110 /* Connection timed out */ #define ECONNREFUSED 111 /* Connection refused */ #define EHOSTDOWN 112 /* Host is down */ #define EHOSTUNREACH 113 /* No route to host */ #define EALREADY 114 /* Operation already in progress */ #define EINPROGRESS 115 /* Operation now in progress */ #define ESTALE 116 /* Stale NFS file handle */ #define EUCLEAN 117 /* Structure needs cleaning */ #define ENOTNAM 118 /* Not a XENIX named type file */ #define ENAVAIL 119 /* No XENIX semaphores available */ #define EISNAM 120 /* Is a named type file */ #define EREMOTEIO 121 /* Remote I/O error */ #define EDQUOT 122 /* Quota exceeded */ #define ENOMEDIUM 123 /* No medium found */ #define EMEDIUMTYPE 124 /* Wrong medium type */ #define ECANCELED 125 /* Operation Canceled */ #define ENOKEY 126 /* Required key not available */ #define EKEYEXPIRED 127 /* Key has expired */ #define EKEYREVOKED 128 /* Key has been revoked */ #define EKEYREJECTED 129 /* Key was rejected by service */ /* for robust mutexes */ #define EOWNERDEAD 130 /* Owner died */ #define ENOTRECOVERABLE 131 /* State not recoverable */ #define ERFKILL 132 /* Operation not possible due to RF-kill */ #endif
就是没有找到139.
A: 看来,系统已经发生一些诡异的情况,错误码已经不正确了。为了确定139错误码确实不存在,我们在/usr/include目录下递归搜索139这个字符。
grep -R '139' *
那么,我们来看看系统日志吧,到底哪里可能有问题。
Q: 使用如下命令得到了错误信息:
最后的地方确实看到hello应用程序运行错误的系统日志。应该是指针访问出错。原因是否是汇编代码大最后没有恰当地设置堆栈寄存器等寄存器的值呢?
A: 在这里,很有可能。为了更容易看出问题可能在哪里,写一个类似功能的c代码,得到它的汇编代码,和上面的汇编代码进行比较。
Q: 写了如下的hello_1.c代码如下:
#include <stdio.h> int main() { printf("Hello,world!\n"); return 0; }查看它的汇编代码:
.file "hello_1.c" .section .rodata .LC0: .string "Hello,world!" .text .globl main .type main, @function main: pushl %ebp movl %esp, %ebp andl $-16, %esp subl $16, %esp movl $.LC0, (%esp) call puts movl $0, %eax leave ret .size main, .-main .ident "GCC: (Ubuntu/Linaro 4.5.2-8ubuntu4) 4.5.2" .section .note.GNU-stack,"",@progbits
A: 不过在实际中,不管是加入pushl %ebp之类代码,还是加入leave, ret指令,最终执行依然是段错误。这个地方笔者一直没明白,如果有谁知道的,希望能不吝赐教。不过,可以调用exit系统调用实现结束应用程序,这样就不会出现段错误。如下:
.section .rodata str: .ascii "Hello,world.\n" .section .text .globl _main _main: movl $4, %eax # the number of system call movl $1, %ebx # file descriptor, 1 means stdout movl $str, %ecx # string address movl $13, %edx # string length int $0x80 movl $1, %eax movl $0, %ebx int $0x80
Q: 进行0x80软中断进行系统调用,参数在哪里保存,就在上面写的寄存器里面吗?
A: 是的。linux下,功能号和返回值在eax中保存,参数一般在5个以下,就按照ebx, ecx, edx, esi, edi来传递,如果参数过多,就会使用堆栈。可以看到上面两次系统调用,均是在使用ebx, ecx, edx这些寄存器。
Q: 4号系统调用是什么?在哪里能知道?
A: 可以在/usr/include/asm/unistd_32.h或者/usr/include/asm/unistd_64.h中看到平台所有系统调用,下面为unistd_32.h文件中开始一部分:
#define __NR_restart_syscall 0 #define __NR_exit 1 #define __NR_fork 2 #define __NR_read 3 #define __NR_write 4 #define __NR_open 5 #define __NR_close 6 #define __NR_waitpid 7 #define __NR_creat 8 #define __NR_link 9 #define __NR_unlink 10 #define __NR_execve 11 #define __NR_chdir 12 #define __NR_time 13 #define __NR_mknod 14 #define __NR_chmod 15 #define __NR_lchown 16 #define __NR_break 17
Q: 汇编如何调用c库函数?
A: 使用call指令,不过调用前要传好参数。如下代码,调用c库printf函数:
.section .rodata str: .ascii "Hello,world.\n" .section .text .globl main main: pushl $str call printf pushl $0 call exit
运行:
Q: 可以使用as, ld来汇编以及链接吗?
A: 可以的。不过需要注意,因为它使用c库,需要指定链接c库: -lc;
Q: 乘法运算mul后面只跟着一个数,另一个数存哪里?
A: 另一个数存储在al, ax或者eax寄存器中,这取决于使用的是mulb, mulw还是mull指令。结果将按照高位到地位的顺序保存在dx和ax中。
同理,除法运算div后面也只跟一个除数,被除数保存在ax, dx:ax或者edx:eax中。除数的最大长度只能是被除数的一半。商和余数将根据被除数占用大小来确定:
如果被除数在ax中,商在al, 余数在ah; 如果被除数在eax中,商在ax, 余数在dx; 如果被除数在edx:eax中,商在eax, 余数在edx.
如下是测试代码:
#include <stdio.h> #include <string.h> #define PRINT_D(longValue) printf(#longValue" is %ld\n", ((long)longValue)); #define PRINT_STR(str) printf(#str" is %s\n", (str)); static void assemble_func() { int result_high, result_low; short result, remainder; // mul __asm__("mov $10, %eax"); __asm__("mov $10, %ebx"); __asm__("mull %ebx"); __asm__("mov %%edx, %0":"=r"(result_high)); __asm__("mov %%eax, %0":"=r"(result_low)); PRINT_D(result_high) PRINT_D(result_low) // div __asm__("mov $0, %dx"); __asm__("mov $100, %ax"); // the divident is dx:ax __asm__("mov $9, %bx"); __asm__("div %bx"); // the divisor is bx __asm__("movw %%ax, %0":"=r"(result)); __asm__("movw %%dx, %0":"=r"(remainder)); PRINT_D(result) PRINT_D(remainder) } int main() { assemble_func(); return 0; }
result_high is 0 result_low is 100 result is 11 remainder is 1
Q: 对于数据比较指令cmp,它是如何配合jmp相关的指令?
A: cmp指令将进行两个数据的差计算,如果得到的是0,jz成立; 如果不是0, jnz成立。如下例子:
#include <stdio.h> #include <string.h> #define PRINT_D(longValue) printf(#longValue" is %ld\n", ((long)longValue)); #define PRINT_STR(str) printf(#str" is %s\n", (str)); #define PRINT(str) printf(#str"\n"); static void assemble_func() { __asm__("mov $10, %eax"); __asm__("cmp $10, %eax "); __asm__("jz end"); PRINT("below jz") __asm__("end:"); PRINT("the end") } int main() { assemble_func(); return 0; }
"the end"
Q: 对于某些时候,加法可能导致溢出,如何判断出来?
A: CPU内部有一个寄存器,它内部会保存溢出标志位OF, 可以通过jo或者jno判断。
#include <stdio.h> #include <string.h> #define PRINT_D(longValue) printf(#longValue" is %ld\n", ((long)longValue)); #define PRINT_STR(str) printf(#str" is %s\n", (str)); #define PRINT(str) printf(#str"\n"); static void assemble_func() { __asm__("movw $0x7FFF, %ax"); __asm__("movw $0x7FFF, %bx"); __asm__("addw %bx, %ax"); __asm__("jo overflow_set"); __asm__("movl $1, %eax"); __asm__("movl $0, %ebx"); __asm__("int $0x80"); __asm__("overflow_set:"); PRINT("overflow flag is set...") } int main() { assemble_func(); return 0; }
"overflow flag is set..."
Q: 对于溢出,到底应该判断?
A: 以加法举例,如果两个相同符号的数相加得到的结果符号相反,那么一定溢出了。
Q: OF和CF标志位有什么区别?
A: CF代表进位标志。进位不一定是溢出,比如有符号整形最小值加1,虽然进位,但是没溢出。因为计算机补码的理论允许进位,但是结果却正确。
#include <stdio.h> #include <string.h> #define PRINT_D(longValue) printf(#longValue" is %ld\n", ((long)longValue)); #define PRINT_STR(str) printf(#str" is %s\n", (str)); #define PRINT(str) printf(#str"\n"); static void assemble_func() { __asm__("movw $0xFFFF, %ax"); __asm__("movw $0x1, %bx"); __asm__("addw %bx, %ax"); __asm__("je carry_set"); __asm__("movl $1, %eax"); __asm__("movl $0, %ebx"); __asm__("int $0x80"); __asm__("carry_set:"); PRINT("carry flag is set...") } int main() { assemble_func(); return 0; }
"carry flag is set..."
#include <stdio.h> #include <string.h> #define PRINT_D(longValue) printf(#longValue" is %ld\n", ((long)longValue)); #define PRINT_STR(str) printf(#str" is %s\n", (str)); #define PRINT(str) printf(#str"\n"); static void assemble_func() { __asm__("movw $0xFFFF, %ax"); __asm__("movw $0x1, %bx"); __asm__("addw %bx, %ax"); __asm__("jo overflow_set"); __asm__("movl $1, %eax"); __asm__("movl $0, %ebx"); __asm__("int $0x80"); __asm__("overflow_set:"); PRINT("overflow flag is set...") } int main() { assemble_func(); return 0; }
执行结果:
它什么也没输出,这就意味着OF没有被置位。
作者:陈曦
日期:2012-6-8 10:50:13
环境:[Ubuntu 11.04 Intel-based x64 gcc4.5.2 CodeBlocks10.05 AT&T汇编 Intel汇编]
转载请注明出处